How would you balance the following equation: Fe + Cl2 + H2O → [Fe(H2O)6]3+ + Cl-?

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How would you balance the following equation: Fe + Cl2 + H2O → [Fe(H2O)6]3+ + Cl-?

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Answer 1 · 2016-01-08T21:47:05

You are oxidizing iron, and then making a coordination complex. I will give you a reaction scheme.

Explanation:

OXIDATION

$F e (s) + \frac{3}{2} C l_{2} (g) \to F e C l_{3} (s)$ $Fe(s) + 3/2Cl_2(g) rarr FeCl_3(s)$

COMPLEXATION

$F e C l_{3} (s) + 6 H_{2} O (l) \to {[F e {(O H_{2})}_{6}]}_{6}^{3 +} {(C l^{-})}_{3}^{-}$ $FeCl_3(s) + 6H_2O(l) rarr [Fe(OH_2)_6]^(3+)(Cl^-)_3$

I don't know whether this is completely what you want.

Answer 2 · 2016-01-21T00:42:29

$2 Fe + {3Cl}_{2} + {12H}_{2} O \to 2 {[Fe {(H_{2} O)}_{6}]}_{6}^{3 +} + {6Cl}^{-}$ $2"Fe" + "3Cl"_2 + "12H"_2"O" → 2["Fe"("H"_2"O")_6]^(3+) + "6Cl"^(-)$

Explanation:

A Socratic answer here shows how to balance redox equations.

Your equation is:

$Fe l + {Cl}_{2} + H_{2} O \to {[Fe {(H_{2} O)}_{6}]}_{6}^{3 +} + {Cl}^{-}$ $"Fe"color(white)(l) + "Cl"_2 + "H"_2"O" → ["Fe"("H"_2"O")_6]^(3+) + "Cl"^(-)$

SOLUTION:

1: The two half-reactions are

$Fe \to {[Fe {(H_{2} O)}_{6}]}_{6}^{3 +}$ $"Fe" → ["Fe"("H"_2"O")_6]^(3+)$
${Cl}_{2} \to {Cl}^{-}$ $"Cl"_2 → "Cl"^(-)$

2: Balance all atoms other than $H$ $"H"$ and $O$ $"O"$ .

$Fe \to {[Fe {(H_{2} O)}_{6}]}_{6}^{3 +}$ $"Fe" → ["Fe"("H"_2"O")_6]^(3+)$
${Cl}_{2} \to 2 {Cl}^{-}$ $"Cl"_2 → color(red)(2)"Cl"^(-)$

3: Balance $O$ $"O"$ .

$Fe l + {6H}_{2} O \to {[Fe {(H_{2} O)}_{6}]}_{6}^{3 +}$ $"Fe"color(white)(l) + color(red)("6H"_2"O") → ["Fe"("H"_2"O")_6]^(3+)$
${Cl}_{2} \to 2 {Cl}^{-}$ $"Cl"_2 → 2"Cl"^(-)$

4: Balance H.

Done.

5: Balance charge.

$Fe l + {6H}_{2} O \to {[Fe {(H_{2} O)}_{6}]}_{6}^{3 +} + {3e}^{-}$ $"Fe"color(white)(l)+ "6H"_2"O" → ["Fe"("H"_2"O")_6]^(3+) + color(red)("3e"^(-))$
${Cl}_{2} + {2e}^{-} \to 2 {Cl}^{-}$ $"Cl"_2 + color(red)("2e"^(-)) → 2"Cl"^(-)$

6: Equalize electrons transferred.

$2 \times {Fe + {6H}_{2} O \to {[Fe {(H_{2} O)}_{6}]}_{6}^{3 +} + + {3e}^{-}}$ $color(red)(2 ×) {"Fe" + "6H"_2"O" → ["Fe"("H"_2"O")_6]^(3+) + + "3e"^(-)}$
$3 \times {{Cl}_{2} + {2e}^{-} \to 2 {Cl}^{-}}$ $color(red)(3 ×){"Cl"_2 + "2e"^(-) → 2"Cl"^(-)}$

7: Add the two half-reactions.

$2 Fe + 3 {Cl}_{2} + {12H}_{2} O \to 2 {[Fe {(H_{2} O)}_{6}]}_{6}^{3 +} + {6Cl}^{-}$ $2"Fe" + 3"Cl"_2 + "12H"_2"O" → 2["Fe"("H"_2"O")_6]^(3+) + "6Cl"^(-)$

8: Check mass balance.

On the left: $2 Fe; 6 Cl; 24 H; 12 O$ $"2 Fe"; "6 Cl"; "24 H"; "12 O"$
On the right: $2 Fe; 24 H; 12 O; 6 Cl$ $"2 Fe"; "24 H"; "12 O"; "6 Cl"$

9: Check charge balance

On the left: $0$ $0$
On the right: $6+ + 6- = 0$ $"6+ + 6-" = 0$

The balanced equation is

$2 Fe + 3 {Cl}_{2} + {12H}_{2} O \to 2 {[Fe {(H_{2} O)}_{6}]}_{6}^{3 +} + {6Cl}^{-}$ $2"Fe" + 3"Cl"_2 + "12H"_2"O" → 2["Fe"("H"_2"O")_6]^(3+) + "6Cl"^(-)$

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How would you balance the following equation: Fe + Cl2 + H2O → [Fe(H2O)6]3+ + Cl-?

2 Answers

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